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Jet001 [13]
4 years ago
12

Use distance formula d=√(X2 – X1 )2 + (X2 – X1 )2, calculate the distance r from the origin to the point r (-2, √5).

Mathematics
1 answer:
lesya692 [45]4 years ago
7 0
d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} \\d = \sqrt{(\sqrt{5} - (-2))^{2} + (\sqrt{5} - (-2))^{2}} \\d = \sqrt{(\sqrt{5} + 2)^{2} + (\sqrt{5} + 2)^{2}} \\d = \sqrt{(4 + 4\sqrt{5} + 5) + (4v + 4\sqrt{5} + 5)} \\d = \sqrt{0 + 0} \\d = \sqrt{0} \\d = 0
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15+3x=45 solve the equation​
inna [77]

Answer:

x= 10

Step-by-step explanation:

45-15=30

30/3=10

3 0
3 years ago
I need help with this a and b please
jarptica [38.1K]
For a just divide a by 46
a/46=c
For b
322=46c
322/46=c
7 circuits =c
8 0
4 years ago
You bought a freezer that is 2 feet wide, 4 feet long and 30 inches deep
Dima020 [189]
2 (w) x 4 (l) x 30 (b) = 240 sq ft.

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7 0
3 years ago
Which lists the steps in the correct order to find the median of this data set? 24, 16, 23, 30, 18, 29 O 1. Put the numbers in o
timofeeve [1]

Median is the central tendency of the data and always exists in the middle of the data set.

<h3>What is the median?</h3>

The Median is the middlemost value of the numbers arranged in the ascending orders. Median is the central tendency of the data and always exists in the middle of the data set.

The process of finding the median if the number of data points is even is as follows:

  • Put the numbers or the data points in the order(increasing or decreasing i.e. ascending or descending order)
  • Next cross of the high/ low pairs
  • Add the leftover numbers
  • and divide the number( as obtained in the above step by 2)

( if the number of data pints were even then we just need to arrange the numbers and cross the high/low pair and only one data point will be left which acts as the Median of the data set)

To know more about median follow

brainly.com/question/26151333

#SPJ1

4 0
2 years ago
Determine whether the pair of lines is parallel, perpendicular, or neither. Y=-6/5x+2 Y=6/5x+2
max2010maxim [7]

k:y=m_1x+b_1;\ l:y=m_2x+b_2\\\\k\ \perp\ l\iff m_1\cdot m_2=-1\\\\k\ ||\ l\iff m_1=m_2

k:y=-\dfrac{6}{5}x+2\to m_1=-\dfrac{6}{5}\\\\l:y=\dfrac{6}{5}x+2\to m_2=\dfrac{6}{5}\\\\m_1\neq m_2\to\text{no parallel}\\\\m_1\cdot m_2=-\dfrac{6}{5}\cdot\dfrac{6}{5}\neq-1\to\text{ no perpendicular}

Answer: NEITHER

4 0
3 years ago
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